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<span>The equation of the linear function is given in the form of a directional
</span><span>equation consists of solid :
1.współczynniki a and b .
2. The variables x and y
</span>y= 3x - 12
y = ax + b
a = 3 (<span>the slope of the tangent to the x-axis)
</span>b = - 12 (<span>intersection of the axis y)
</span>we look for the point ( 0 , -12 ) because b = -12
A. (4, 0) ,(0, -12)
B.(12 ,0), (0,4)
C.(3, 0), (0, 4)
D.(4, 0), (0, 3)
E.(-4, 0), (0, -12)
<span>fit and answers a,e
</span>
<span>we set the coordinate system and you activate the point ( 0 , -12 )
</span><span>Now draw a straight line parallel to the x axis przchodzącą through the point (0 , -12 )
</span><span>We notice that a> 0 the function is increasing
</span><span>We draw a line parallel to the triangle . Invent lengths of the sides 3 : 1. Examples are a = b = 1cm 3cm and 6cm a = b = 2 cm .
</span><span>the hypotenuse of the triangle is our simple
</span><span>Intersection of the axis x = ( 4,0 )
Answer A</span>
</span><span>equation consists of solid :
1.współczynniki a and b .
2. The variables x and y
</span>y= 3x - 12
y = ax + b
a = 3 (<span>the slope of the tangent to the x-axis)
</span>b = - 12 (<span>intersection of the axis y)
</span>we look for the point ( 0 , -12 ) because b = -12
A. (4, 0) ,(0, -12)
B.(12 ,0), (0,4)
C.(3, 0), (0, 4)
D.(4, 0), (0, 3)
E.(-4, 0), (0, -12)
<span>fit and answers a,e
</span>
<span>we set the coordinate system and you activate the point ( 0 , -12 )
</span><span>Now draw a straight line parallel to the x axis przchodzącą through the point (0 , -12 )
</span><span>We notice that a> 0 the function is increasing
</span><span>We draw a line parallel to the triangle . Invent lengths of the sides 3 : 1. Examples are a = b = 1cm 3cm and 6cm a = b = 2 cm .
</span><span>the hypotenuse of the triangle is our simple
</span><span>Intersection of the axis x = ( 4,0 )
Answer A</span>
Answer:
See attached picture and description below.
Step-by-step explanation:
Graph each equation as you would graph any linear function. Plot the y-intercept for the first equation at (0,5) then move down 0.5 units and to the right 1 unit. This is the right boundary and should be plotted with an open circle. (red graph.
The second equation should be graphed starting at the y-intercept (0,-2.5) then proceeding down 0.5 units and over 1 unit to the right. (blue graph)
See graph below for complete graph.
Answer:
Its 2πr^2
Step-by-step explanation: